TITLE: Sipser's proof of NP-completeness of SAT 
QUESTION [6 upvotes]: I just need some help with the Cook-Levin theorem (SAT is NP-complete).
The proof of this theorem is not so clear to me.
I use Sipser's book "Introduction to the Theory of Computation". 
Up to this proof, everything else seems very clear and obvious; therefore I chose this book. The proof is on the page 277.
I can't understand the concept of the tableau from the very beggining .
In particular,

every row in tablea presents branch with states if corresponding nodes in TM?
what is the idea of window stands for ?

May be there is any simplified version, you know of?
Thanks!

Thanks for great explanation!
I would like to ask you few more questions
the idea of tableau is clear
the idea of cell is clear 
but I still have a problem with designing $\phi$ (correspond boolean expression to input $\omega$)
Now we design $\phi$ so that a satisfing assignment to the variables does correspond to an accepting tableau for N on $\omega$. The formula for $\phi$ is the AND of four parts $\phi_{cell}\wedge\phi_{start}\wedge\phi_{move}\wedge\phi_{accept} $ (how can this construction describe $\phi$?) . We describe each part in turn.
As we mentioned previously, tunrning variable $x_{i,j,s}$ on corresponds to placing symbol $s$ in $cell[i,j]$  The first thing we must guarantee in order to obtain a correspondence beetwen  an assignment and a tableau is that the assignment turns on exactly one variable for each cell. Formula $\phi_{cell}$ ensures this requirments by expressing it in terms of Boolean operations
$$\phi_{cell} = \underset{1\leqslant i, j\leqslant n^{k}}{\wedge }\left [  \left (  \underset {s\epsilon C}{\wedge} x_{x,j,s} \right ) \wedge \left (\underset{\underset{s\neq t}{s,t\epsilon C}}{\wedge}  \left ( \overline{x_{i,j,s}} \vee \overline{x_{i,j,t}} \right ) \right ) \right ]$$
Why $\phi_{cell}$ has a above equation?
Thanks!

REPLY [3 votes]: I'm using a second answer to answer your newly added question. You made a little mistake, the cell formula should be the following: 
$$\phi_{cell} = \underset{1\leqslant i, j\leqslant n^{k}}{\wedge }\left [  \left (  \underset {s\epsilon C}{\vee} x_{i,j,s} \right ) \wedge \left (\underset{\underset{s\neq t}{s,t\epsilon C}}{\wedge}  \left ( \overline{x_{i,j,s}} \vee \overline{x_{i,j,t}} \right ) \right ) \right ]$$
As you said, the variable $x_{i,j,s}$ represents the symbol s in cell[i,j], so it should be true if and only iff cell[i,j] contains the symbol $s$. Let's analyse the cell formula properly. It exists out of 3 parts:

$\Large\underset {s\epsilon C}{\vee} x_{i,j,s}$ tells that for each fixed $i$ and $j$ there should be atleast one symbol $s$ in cell[i,j].
$\large\underset{\underset{s\neq t}{s,t\epsilon C}}{\wedge}  \left ( \overline{x_{i,j,s}} \vee \overline{x_{i,j,t}} \right )$ says that, for each distinct pair of symbols $s$ and $t$: $\large x_{i,j,s}$ and $\large x_{i,j,t}$ cannot both be true at the same time. This represents that a cell cannot have more then one symbol.

Combining both points tells us that for each fixed $i$ and $j$ there has to be exactly one symbol $s$ ( NOT MORE AND NOT LESS) so that variable $x_{i,j,s} = 1$. More intuitively it tells us that each cell can only contain exactly one symbol. 
Now the only thing left is to analyse is the $\large\underset{1\leqslant i, j\leqslant n^{k}}{\wedge }$ part encapsulating the parts we already analysed. Clearly this part just tells us that the formula has to hold for each cell in the tableau.
I hope this helped.